Question

Derivative trick Here is an alternative way to evaluate higher derivatives of a function f that may save time. Suppose you can find the Taylor series for f centered at the point a without evaluating derivatives (for example, from a known series). Then f⁽ᵏ⁾(a)=k! multiplied by the coefficient of (x−a)ᵏ. Use this idea to evaluate f⁽³⁾(0) and f⁽⁴⁾(0) for the following functions. Use known series and do not evaluate derivatives.

f(x) = ∫₀ˣ sin t² dt

Accepted Answer

Recognize that the function is given by an integral: $$f(x) = \$$int_0$$^x \sin(t^2$) \, dt. To $$find the higher derivatives at 0 without directly differentiating, we will use the Taylor series expansion of $$\sin(t^2$)$$ around 0. Recall the Taylor series for $$\sin z$$ centered at 0: $$\sin z = \sum_{n=0}^\infty (-1)^n \frac{z$$^{2n+1}$$}{(2n+1)!}. $$Substitute $$z = t^2$ to get the series for $$\$$sin(t^2$$)$$: $$\$$sin(t^2$$) = \sum_{n=0}^\infty (-1)^n \frac{$$t^{4n+2}$$}{(2n+1)!}$$. Integrate the series term-by-term from 0 to x$ to find the series for f(x$$)$$: f(x$$) = \$$int_0^x$$ \$$sin(t^2$$) \, dt = \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!} \$$int_0^x$$ $$t^{4n+2}$$ \, dt$$. Perform the integral inside the sum to get f(x$$) = \sum_{n=0}^\infty (-1)^n \frac{$$x^{4n+3}$$}{(2n+1)! (4n+3)}$$. Identify the coefficients of the powers of x$ in the series expansion of f(x$$)$$: the general term is a_n$$ $$x^{4n+3}$$ where $$a_n = (-1)^n / [(2n+1)! (4n+3)]. $$Since the series is centered at 0, the coefficient of $$x^k$$ corresponds to the coefficient of $$(x-0)^k. $$Use the formula for the $$k-th $$derivative at 0: $$f^{(k)}(0) = k$$!$$ times the coefficient of x^k$ in the Taylor series. To find f$$^{(3)}$$(0)$$ and $$f^{(4)}(0$$)$$, locate the coefficients of x^3$ and x^4$ respectively in the series and multiply each by 3$$!$$ and 4$$!$$. Note that if a power does not appear in the series, the corresponding derivative at 0 is zero.

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Key Concepts

Taylor Series Expansion

Relationship Between Taylor Coefficients and Derivatives

Using Known Series to Find Derivatives of Integral Functions